69.5pointseBookCalculate the standard error. May normality be assumed? (Round your answers to 4 decimal places.)(a)(b)(c)(d)n = 27, TT= 0.32n = 58, TT= 0.45n = 119, π = 0.39n = 359, TT = 0.004Standard ErrorNormalityNoNoYesNo 2For each value of d.f. look up the value of Student's t in Appendix D for the stated level of confidence. How close is the t-value to thecorresponding z-value (at the bottom of the column for d.f. = 00)?Use Appendix D and Excel to find the value of Student's t for the given level of confidence. (Round your answers to 4 decimal places.)(a-1) d.f. 40, 95% confidence.9.5pointseBookPrintReferencesThe value of Student's t(a-2) The z-value is close.O YesO No(b-1) d.f. 80, 95% confidence.The value of Student's t(b-2) The z-value is close.● YesO No(c-1) d.f. = 100, 95% confidence.The value of Student's t(c-2) The z-value is close.● YesO No

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question Help Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of the table. Click to view page 2 of the table. 0.7 The indicated IQ score, x, is (Round to one decimal place as needed.) 21
In a study to investigate the extent to which individuals are aware of industrial odors in a certain regiont, a sample of individuals was obtained from each of three different areas near industrial facilities. Each individual was asked whether he or she noticed odors (1) every day, (2) at least once/week, (3) at least once/month, (4) less often than once/month, or (5) not at all, resulting in the data and SPSS output below. SPSS output Crosstabulation: AREA BY CATEGORY Count Val Exp CATEGORY → Row Pct Row AREA Col Pct 1.00 2.00 3.00 4.00 5.00 Total 1.00 20 28 23 14 12 97 12.7 24.7 18.0 16.0 25.7 33.38 20.6% 28.9% 23.78 14.48 12.4% 52.6% 37.8% 42.6% 29.28 15.6% 2.00 14 34 21 14 12 95 12.4 24.2 17.6 15.7 25.1 32.6% 14.78 35.88 22.1% 14.78 12.6% 36.88 45.9% 38.9% 29.28 15.6% 3.00 4 12 10 20 53 99 12.9 25.2 18.4 16.3 26.2 34.0% 4.08 12.1% 10.18 20.2% 53.5% 10.5% 16.2% 18.5% 41.78 68.88 Column 38 74 54 48 77 291 Total 13.18 25.4% 18.6% 16.5% 26.5% 100.0% Chi-Square D.F. Significance Min…
C ī Biostatistics What formula would you use to find A point estimate for the mean change in heart rate (level I to baseline?? in
Find the value of the test statistic. T* = 51.50 Find the p-value. (Round your answer to four decimal places.) p-value = 0.0071 X
The authors of the paper "Evaluating Existing Movement Hypotheses in Linear Systems Using Larval Stream Salamanders"+ investigated whether water temperature was related to how far a salamander would swim and whether it would swim upstream or downstream. Data for 14 streams with different mean water temperatures where salamander larvae were released are given (approximated from a graph that appeared in the paper). The two variables of interest are x = mean water temperature (°C) and y = net directionality, which was defined as the difference in the relative frequency of the released salamander larvae moving upstream and the relative frequency of released salamander larvae moving downstream. A positive value of net directionality means a higher proportion were moving upstream than downstream. A negative value of net directionality means a higher proportion were moving downstream than upstream. Mean Temperature (x) Net Directionality (y) 6.17 -0.08 8.06 0.25 8.62 -0.14 10.56 0.00 12.45…
Suppose patients are graded on the degree of change in periodontal status on a 7-point scale, with +3 indicating the greatest improvement, 0 indicating no change, and −3 indicating the greatest decline. The data are given in the table below. Degree of Change in Periodontal Status Change Score Number of Patients +3 +2 +1 0 -1 -2 -3 3 4 5 4 5 2 2 (b) Taking into account the 7-point scale, what nonparametric test can be used to determine whether a significant change in periodontal status has occurred over time? the sign test the Wilcoxon rank-sum test the Wilcoxon signed-rank test
plz solve both parts within 30-40 mins I'll upvote your answer
IQ tests supposedly measure our intelligence. They typically are ‘normed’ so that they have a mean of 100 and a standard deviation of 15. (Higher scores show higher intelligence.) Use the graph and use the z-score and percents from columns b and c in Table A, when relevant. If you scored 132, what is your z-score? (Interpret.) Use table A X, z(a), (b), and (c) X z (a) (b) (c) and use the graph! Write down or save your response to use in Part 2 of this quiz. Group of answer choices -2.53 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. 2.13 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. -2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 0.57 percentile rank and 99.43% of people scored smarter than me. 2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 99.43 percentile rank…
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